Problem: Given a positive integer $n$, it can be shown that every complex number of the form $r+si$, where $r$ and $s$ are integers, can be uniquely expressed in the "base" $-n+i$ using the integers $1,2,\ldots,n^2$ as "digits." That is, the equation\[r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0\]is true for a unique choice of non-negative integer $m$ and digits $a_0,a_1,\ldots,a_m$ chosen from the set $\{0,1,2,\ldots,n^2\}$, with $a_m\ne 0$. We write \[r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}\]to denote the base $-n+i$ expansion of $r+si$.

There are only finitely many integers $k+0i$ that have four-digit expansions
\[k=(a_3a_2a_1a_0)_{-3+i} \qquad (a_3 \neq 0).\]Find the sum of all such $k$.
To say that $k = (a_3a_2a_1a_0)_{-3+i}$ is to say that \[k = a_3(-3+i)^3 + a_2(-3+i)^2 + a_1(-3+i) + a_0.\]Expanding the right-hand side, we have  \[k = (-18a_3+8a_2-3a_1+a_0) + (26a_3-6a_2+a_1)i.\]Since $k$ is a real number, the imaginary part of the right-hand side must be zero; that is, \[26a_3 - 6a_2 + a_1 = 0\]or \[26a_3 = 6a_2 - a_1.\]Remember that $0 \le a_1, a_2, a_3\le 9$, so $6a_2 - a_1 \le 6 \cdot 9 - 0 = 54$. Thus, $26a_3 \le 54$, so $a_3 \le 2$. We take cases, remembering that $a_3 \neq 0$:

If $a_3 = 1$, then we have $6a_2 - a_1 = 26$. The only solution to this equation is $(a_1, a_2) = (4, 5)$, so we have \[k = -18a_3 + 8a_2 - 3a_1 + a_0 = -18 \cdot 1 + 8 \cdot 5 -3 \cdot 4 + a_0 = 10 + a_0.\]Since $a_0 \in \{0, 1, 2, \ldots, 9\}$, the possible values of $k$ are $10, 11, 12, \ldots, 19$, and these have a sum \[10 + 11 + 12 + \dots + 19 = \frac{29 \cdot 10}{2} = 145.\]
If $a_3 = 2$, then we have $6a_2 - a_1 = 52$. The only solution to this equation is $(a_1, a_2) = (2, 9)$, so we have \[k = -18a_3 + 8a_2 - 3a_1 + a_0 = -18 \cdot 2 + 8 \cdot 9 -3 \cdot 2 + a_0 = 30 + a_0.\]Therefore, the possible values of $k$ are $30, 31, 32, \ldots, 39$, which sum to \[30 + 31 + 32 + \dots + 39 = \frac{69 \cdot 10}{2} = 345.\]

Adding up both cases, we get the answer, $145 + 345 = \boxed{490}$.